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  1. Prepare
  2. Data Structures
  3. Linked Lists
  4. Get Node Value
  5. Discussions

Get Node Value

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982 Discussions

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  • nizmu
     + 0 comments

    My Python Soluction

    def getNode(llist, positionFromTail):

        prev = None
        curr = llist
        temp = curr

        while curr:
                temp = curr.next
                curr.next = prev
                prev = curr
                curr = temp

        head = prev
        curr = head
        count = 0

        while count < positionFromTail:
                curr = curr.next
                count += 1

        return curr.data

  • mdsanz
     + 0 comments

    My Java 8 Solution

    public static int getNode(SinglyLinkedListNode llist, int positionFromTail) {
        int lenght = 0;
        SinglyLinkedListNode counterList = llist;
        
        while (counterList != null) {
            lenght++;
            counterList = counterList.next;
        }
        
        int positionFromHead = lenght - positionFromTail - 1;
        int index = 0;
        SinglyLinkedListNode node = llist;
        
        while (index < positionFromHead) {
            node = node.next;
            index++;
        }
        
        return node.data;
    }

  • rajpatel92111213
     + 0 comments
    public static int getNode(SinglyLinkedListNode head, int positionFromTail) {
    // Create two pointers, both starting at the head
    SinglyLinkedListNode fast = head;
    SinglyLinkedListNode slow = head;

    // Move the 'fast' pointer 'positionFromTail' steps ahead
    for (int i = 0; i < positionFromTail; i++) {
        fast = fast.next;
    }

    // Now move both 'fast' and 'slow' one step at a time
    // When 'fast' reaches the last node, 'slow' will be at the target node
    while (fast.next != null) {
        fast = fast.next;
        slow = slow.next;
    }

    // 'slow' is now pointing to the node that is 'positionFromTail' from the end
    return slow.data;
}

  • deepu140698
     + 0 comments

    def getNode(llist, positionFromTail): cur = llist stack = [] while cur : stack.append(cur.data) cur = cur.next for i in range(positionFromTail+1): node_value = stack.pop() return node_value

  • getakhilesh705
     + 0 comments

    simple recursive solution in linear time

        static int count = 0;
    static int val = 0;
    public static int getNode(SinglyLinkedListNode llist, int positionFromTail) {
    if(llist==null){
        count=0;
        return 0;
    }
    getNode(llist.next, positionFromTail);
    if(positionFromTail==count){
        val=llist.data;
    }
    count++;
    return val;
    }